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Triangles — Maths STD 10 — Question

CBSE BoardEnglish MediumSTD 10MathsTriangles5 Marks
Question
In the given figure, $A B C$ and $D B C$ are two triangles on the same base $B C$. If $A D$ intersects $B C$ at $O$, show that $\frac{\operatorname{ar}(\triangle A B C)}{\operatorname{ar}(\triangle D B C)}=\frac{A O}{D O}$
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Answer

Given: $B C$ is the common base of $\triangle A B C$ and $(\triangle D B C)$.
To Prove: $\frac{\operatorname{ar}(\triangle A B C)}{\operatorname{ar}(\triangle D B C)}=\frac{A O}{D O}$
Draw,
$A E \perp B C$ and $D F \perp B C$ in $\triangle A B C$ and $\triangle D B C$ respectively.

Image
Hence, $\operatorname{ar}(\triangle A B C)=\frac{1}{2} \times B C \times A E\quad \quad \ldots \ldots\text{(i)}$
And, $\operatorname{ar}(\triangle D B C)=\frac{1}{2} \times B C \times D F\quad \quad \ldots \ldots\text{(ii)}$
Taking ratio of equation (i) and equation (ii), we get,
$\frac{\operatorname{ar}(\triangle A B C)}{\operatorname{ar}(\triangle D B C)}=\frac{A E}{D F}$
But we need to prove that,
$\frac{\operatorname{ar}(\triangle A B C)}{\operatorname{ar}(\triangle D B C)}=\frac{A O}{D O}$
Hence, we need to prove that, $\frac{A O}{D O}=\frac{A E}{D F}$
In $\triangle A O E$ and $\triangle D O F$,
$\angle A E O=\angle D F O=90^{\circ}$
$\angle A O E=\angle D O F \quad$ (Vertically opposite angles)
Hence, by $A A$ similarity criterion
$\triangle A O E=\triangle D O F$
We know that, if two triangles are similar, their corresponding sides are in the same ratio
Therefore, $\frac{A E}{D F}=\frac{A O}{D O}\quad \quad \ldots \ldots\text{(iii)}$
Now, we have,
$\frac{\operatorname{ar}(\triangle A B C)}{\operatorname{ar}(\triangle D B C)}=\frac{A E}{D F}\quad \quad \ldots \ldots\text{(iv)}$
On comparing equation (iii) and equation (iv), we get,
$\frac{a r(\triangle A B C)}{a r(\triangle D B C)}=\frac{A O}{D O}$
Hence proved.

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